Tank X initially contains 50 liters of a solvent solution that is 20\% chemical compound b… — GMAT

Tank X initially contains $50$ liters of a solvent solution that is $20\%$ chemical compound by volume. Tank Y initially contains $30$ liters of a solvent solution that is $70\%$ of the same chemical compound by volume. First, $v$ liters of the solution in Tank X are transferred to Tank Y and thoroughly mixed. Then, $v$ liters of the resulting mixture in Tank Y are transferred back to Tank X. If the final concentration of the chemical compound in Tank X is $32\%$ by volume, what is the value of $v$ ?

Cevap: 20 liters

Cevap

The value of

v

is 20.

By setting up the conservation of solute equation for Tank X, the initial

10

liters of compound minus the

0.20v

liters transferred to Tank Y, plus the return portion

v \cdot \frac{21 + 0.20v}{30 + v}

equals the final

16

liters of compound. Solving this relation yields

v = 20

Adım Adım Çözüm

Calculate the initial volume of pure chemical compound in each tank.

Tank X initially contains

50 \times 0.20 = 10

liters of compound. Tank Y initially contains

30 \times 0.70 = 21

liters of compound.

Establishing initial solute quantities is essential for building the mass balance equations.

Determine the concentration of Tank Y after the first transfer of

v

liters from Tank X.

Tank Y contains

30 + v

total liters of solution and

21 + 0.20v

liters of compound. Its concentration becomes

C_Y = \frac{21 + 0.20v}{30 + v}

The solution transferred from Tank X carries a 20% concentration into Tank Y, altering Tank Y's volume and concentration.

Formulate the total compound equation for Tank X after transferring

v

liters back from Tank Y.

Remaining compound in X after first transfer =

10 - 0.20v

. Compound returned from Y =

v \cdot C_Y = v \cdot \frac{21 + 0.20v}{30 + v}

. Final compound in X =

50 \times 0.32 = 16

liters.

Tank X ends with its original total volume (50 liters) at a 32% concentration.

Solve the algebraic equation for

v

(10 - 0.20v) + \frac{21v + 0.20v^2}{30 + v} = 16

-0.20v + \frac{21v + 0.20v^2}{30 + v} = 6

Multiply by

(30 + v)

-0.20v(30 + v) + 21v + 0.20v^2 = 6(30 + v)

-6v - 0.20v^2 + 21v + 0.20v^2 = 180 + 6v

15v = 180 + 6v \implies 9v = 180 \implies v = 20

The quadratic terms cancel out cleanly, leaving a simple linear equation.

Anahtar Kavram

Two-Stage Transfer and Replacement Mixture Balance

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